A graph plotted between log (t) 50% vs. log (a) concentration is a straight line. What conclusion can you draw from the given graph?

 

1. n=1, t_1/2 = 1/K.a

2. n=2, t_1/2 = 1/a

3. n=1, t_1/2 = 0.693/K

4. None of the above

The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about :

1. 10 times

2. 24 times

3. 32 times

4. 64 times

Select the intermediate in the following reaction mechanism:

O₃(g) $\rightleftharpoons$ O₂(g) +O(g)

O(g) +O₃(g) $\to$ 2O₂(g)

1. O₃(g)

2. O(g)

3. O₂(g)

4. none of these

For the reaction:

[Cu(NH₃)₄]²⁺ + H₂O_{$\rightleftharpoons$}[Cu(NH₃)₃H₂O]²⁺ + NH₃

the net rate of reaction at any time is given by, net rate =

2.0x10^-4 [Cu(NH₃)₄]²⁺[H₂O] - 3.0x10⁵ [Cu(NH₃ )₃ H₂0]²⁺[NH₃]

Then correct statement is/are :

1. rate constant for forward reaction = 2 x 10^-4

2. rate constant for backward reaction = 3 x 10⁵

3. equilibrium constant for the reaction = 6.6 x 10^-10

4. all of the above

How much faster would a reaction proceed at 25°C than at 0°C if the activation energy is 65 kJ?

1. 2 times

2. 16 times

3. 11 times

4. 6 times

In the Arrhenius equation K = Ae^-E_a/RT, the quantity e^-E_a/kT is referred as:

1. Boltzmann factor.

2. Frequency factor.

3. Activation factor.

4. None of the above.

In gaseous reactions important for the understanding of the upper atmosphere H₂O and O react bimolecularly to form two OH radicals. $∆$H for this reaction is 72kJ at 500 K and E_a is 77 kJ mol^-1, then E_a for the bimolecular recombination of two OH radicals to form H₂O and O is:

1. 3 kJ mol^-1

2. 4 kJ mol^-1

3. 5 kJ mol^-1

4. 7 kJ mol^-1

The activation energies of the forward and backward reactions in the case of a chemical reaction are 30.5 and 45.4 KJ/mol respectively. The reaction is

1. Exothermic

2. Endothermic

3. Neither exothermic nor endothermic

4. Independent of temperature

The correct expression for the 3/4th life of a first-order reaction is:

$1. \frac{\mathrm{k}}{2.303}\log \frac{4}{3}$
$2. \frac{2.303}{\mathrm{k}}\log \frac{3}{4}$
$3. \frac{2.303}{\mathrm{k}}\log$ $4$
$4. \frac{2.303}{\mathrm{k}}$ $\log$ $3$

The dissociation of H₂O₂ is a first order reaction. The half life for '16 V H₂O₂ is 30 min, calculate the time at which the solution is 1V H₂O₂ ?

(1) 120 min.

(2) 90 min.

(3) 60 min.

(4) 150 min.